Triangles Test

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Triangles Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

In figure, it $$DE\parallel BC$$, then $$x$$ will be:

A
$$\sqrt 5$$

B
$$\sqrt 6$$

C
$$\sqrt 3$$

D
$$\sqrt 7$$

Solution

In $$\triangle ABC$$
$$DE\parallel BC$$
$$\cfrac{AD}{BD}=\cfrac{AE}{CE}$$ (Basic Proportionality)
$$\Rightarrow$$ $$\cfrac{x+4}{x+3}=\cfrac{2x-1}{x+1}$$ (By cross multiplication method)
$$\Rightarrow$$ $$(x+4)(x+1)=(x+3)(2x-1)$$
$$\Rightarrow$$ $${x}^{2}+4x+x+4=2{x}^{2}-x+6x-3$$
$$\Rightarrow$$ $${x}^{2}+5x+4=2{x}^{2}+5x-3$$
$$\Rightarrow$$ $${x}^{2}=7$$
$$\Rightarrow$$ $$x=\pm \sqrt{7}$$
Negative value of $$x$$ is not possible
$$x=\sqrt{7}$$
Question 2
1 / -0

In $$\triangle ABC$$ and $$\triangle DEF$$ if $$\cfrac{AB}{DE}=\cfrac{BC}{FD}$$ then they will be similar if:

A
$$\angle B=\angle E$$

B
$$\angle A=\angle D$$

C
$$\angle B=\angle D$$

D
$$\angle A=\angle F$$

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$ , if $$\cfrac{AB}{DE}=\cfrac{BC}{FD}$$ then they will be similar by SAS if the angles between the pair of corresponding sides is equal
i.e $$\angle B =\angle D$$
Question 3
1 / -0

If one angle of a triangle is equal to the sum of the other two angles then the triangle is:

A
Isosceles triangle

B
Obtuse angled triangle

C
Equilateral triangle

D
Right-angled triangle

Solution
Question 4
1 / -0

In $$\triangle ABC$$, given below, $$AB= 8\text{ cm}$$, $$BC= 10\text{ cm}$$ and $$AC= 6\text{ cm}$$. If a point $$P$$ lies on $$AB$$ and $$Q$$ on $$AC$$ such that $$PQ \parallel BC$$ and $$PQ = 5\text{ cm}$$, then find $$AP$$.

A
$$4\text{ cm}$$

B
$$5\text{ cm}$$

C
$$3\text{ cm}$$

D
$$2\text{ cm}$$

Solution

In $$\triangle APQ$$ and $$\triangle ABC$$,
line $$PQ \parallel$$ side $$BC$$ ...given
$$\therefore \angle APQ \cong \angle PBC$$ ...corresponding angles
and $$\angle AQP \cong \angle QCB$$ ...corresponding angles
Also, $$\angle PAQ \cong \angle BAC$$ ...common angle
$$\therefore \triangle APQ \sim \triangle ABC$$ .... by AAA test of similarity

$$\therefore \dfrac {AP}{AB} = \dfrac {AQ}{AC} = \dfrac {PQ}{BC}$$ ( In a pair of similar triangles, the corresponding sides are proportional )

$$\therefore \dfrac {AP}{AB} = \dfrac {AQ}{AC} = \dfrac {5}{10} = \dfrac 12$$

$$\therefore AP = \dfrac 12 AB$$ and $$AQ = \dfrac 12 AC$$
.
$$\therefore AP = \dfrac 12 \times 8$$ and $$AQ = \dfrac 12 \times 6$$

$$\therefore AP = 4\text{ cm}$$ and $$AQ = 3\text{ cm}$$

So, option A is correct.
Question 5
1 / -0

In triangle ABC, AD$$=$$DB, DE is parallel to BC, and the area of triangle ABC is $$40$$. What is the area of triangle ADE?

A
$$10$$

B
$$15$$

C
$$20$$

D
$$30$$

E
It cannot be determined from the information given

Solution

Question 6

1 / -0

Match the column.

1. In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta PQR$$, $$\displaystyle \frac{AB}{PQ}=\frac{AC}{PR},\angle A=\angle P$$	(a) AA similarity criterion
2. In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta PQR$$, $$\displaystyle \angle A=\angle P,\angle B=\angle Q$$	(b) SAS similarity criterion
3. In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta PQR$$, $$\displaystyle \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$$ $$\angle A=\angle P$$	(c) SSS similarity criterion
4. In $$\displaystyle \Delta ACB,DE\|\|BC$$ $$\displaystyle \Rightarrow \frac{AD}{BD}=\frac{AE}{CE}$$	(d) BPT

$$\displaystyle 1\rightarrow a,2\rightarrow b,3\rightarrow c,4\rightarrow d$$

$$\displaystyle a\rightarrow d,2\rightarrow a,3\rightarrow c,4\rightarrow b$$

$$\displaystyle 1\rightarrow b,2\rightarrow a,3\rightarrow c,4\rightarrow d$$

$$\displaystyle 1\rightarrow c,2\rightarrow b,3\rightarrow d,4\rightarrow a$$

Solution

In $$\triangle ABC$$ and $$\triangle PQR$$

Option A:

If $$\angle A = \angle P$$ ....Given

And, $$\dfrac {AB}{PQ} = \dfrac {AC}{PR}$$ ...Given

$$\triangle ABC \sim \triangle PQR$$ ...SAS test of similarity

Option B:

If $$\angle A = \angle P$$ ....Given

And $$\angle B = \angle Q$$ ....Given

$$\triangle ABC \sim \triangle PQR$$ ...AA test of similarity

Option C:

If $$\angle A = \angle P$$ ....Given

And $$\dfrac {AB}{PQ} = \dfrac {AC}{PR} = \dfrac {BC}{QR}$$ ....Given

$$\triangle ABC \sim \triangle PQR$$ ...SS S test of similarity

Option D:

In $$\triangle ACB, DE \parallel BC$$

$$\dfrac {AD}{BD} = \dfrac {AE}{CE} $$ ....Given

This is known as basic proportionality theorem.

Question 7
1 / -0

PQRS is a parallelogram and ST$$=$$TR. What is the ratio of the area of triangle $$QST$$ to the area of the parallelogram?

A
$$1:2$$

B
$$1:3$$

C
$$1:4$$

D
$$1:5$$

E
It cannot be determined

Solution

Given that $$ST=TR$$ So, T is the midpoint of SR.
Now, we know that a diagonal divider the parallelogram into two triangles of equal area.
So, Area of $$\triangle PQS=Area \, of \triangle QRS=\dfrac{1}{2}$$ [area of parallelogram PQRS] -----------(1)
Now, in $$\triangle QRS,QT$$ divides it into two $$\triangle$$. Also height of $$\triangle QST= height \, of \triangle QTR.$$
So, Area of $$\triangle QST=Area \, of \triangle QTR=\dfrac{1}{2}[area \, of \triangle QRS]$$ -----------(2)
from (1),
Area of $$\triangle QRS=\dfrac{1}{2}$$ [area of parallelogram PQRS] ---------(3)
From (2) Area of $$QST=\dfrac{1}{2}[area \, of \triangle QRS]$$
From (2),
Area of $$\triangle QST=\dfrac{1}{2}[area \, of \triangle QRS]$$
$$\Rightarrow $$ Area of $$\triangle QRS=2[area \, of \triangle QST]$$ --------(4)
LHS of (3) and (4) are some so , equating the RHS of (3) and (4) we get.
$$\dfrac{1}{2}[area of parallelogram PQRS]$$ $$=2$$ [area of $$\triangle QST$$]
$$\Rightarrow $$ Area of parallelogram $$PQRS=4$$ (area of \triangle QST)
$$\Rightarrow \dfrac{Area \, of \triangle QST}{Area \, of \, parallelogram PQRS}=\dfrac{1}{4}$$
$$\therefore $$ The required ratio is $$1:4$$
Question 8
1 / -0

Let $$WXYZ$$ be a square. Let $$P,Q,R$$ be the mid points of $$WX, XY$$ and $$ZW$$ respectively and $$K,L$$ be the mid-points of $$PQ$$ and $$PR$$ respectively. What is the value of $$\cfrac { area\quad of\quad triangle\quad PKL }{ area\quad of\quad square\quad WXYZ } $$?

A
$$\cfrac { 1 }{ 32 } $$

B
$$\cfrac { 1 }{ 16 } $$

C
$$\cfrac { 1 }{ 8 } $$

D
$$\cfrac { 1 }{ 64 } $$
Question 9
1 / -0

Three circular laminas of the same radius are cut out from a larger circular lamina. When the radius of each lamina cut out is the largest possible, then what is the ratio (approximate) of the area of the residual piece of the original lamina to its original total area?

A
$$0.30$$

B
$$0.35$$

C
$$0.40$$

D
$$0.45$$

Solution
Question 10
1 / -0

In $$\Delta ABC,\,\,if\,\,D\,\,and\,\,E$$ are mid points of BC and AD respectively such that $$ar\left( {AEC} \right) = 4c{m^2}$$ then $$ar\left( {BEC} \right) = $$

A
$$4c{m^2}$$

B
$$6c{m^2}$$

C
$$8c{m^2}$$

D
$$12c{m^2}$$

Solution

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Re-Attempt Weekly Quiz Competition

Triangles Test - 42

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Triangles Test - 42

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Question 1

$$\sqrt 5$$

$$\sqrt 6$$

$$\sqrt 3$$

$$\sqrt 7$$

Solution

Question 2

$$\angle B=\angle E$$

$$\angle A=\angle D$$

$$\angle B=\angle D$$

$$\angle A=\angle F$$

Solution

Question 3

Isosceles triangle

Obtuse angled triangle

Equilateral triangle

Right-angled triangle

Solution

Question 4

$$4\text{ cm}$$

$$5\text{ cm}$$

$$3\text{ cm}$$

$$2\text{ cm}$$

Solution

Question 5

$$10$$

$$15$$

$$20$$

$$30$$

It cannot be determined from the information given

Solution

Question 6

$$\displaystyle 1\rightarrow a,2\rightarrow b,3\rightarrow c,4\rightarrow d$$

$$\displaystyle a\rightarrow d,2\rightarrow a,3\rightarrow c,4\rightarrow b$$

$$\displaystyle 1\rightarrow b,2\rightarrow a,3\rightarrow c,4\rightarrow d$$

$$\displaystyle 1\rightarrow c,2\rightarrow b,3\rightarrow d,4\rightarrow a$$

Solution

Question 7

$$1:2$$

$$1:3$$

$$1:4$$

$$1:5$$

It cannot be determined

Solution

Question 8

$$\cfrac { 1 }{ 32 } $$

$$\cfrac { 1 }{ 16 } $$

$$\cfrac { 1 }{ 8 } $$

$$\cfrac { 1 }{ 64 } $$

Question 9

$$0.30$$

$$0.35$$

$$0.40$$

$$0.45$$

Solution

Question 10

$$4c{m^2}$$

$$6c{m^2}$$

$$8c{m^2}$$

$$12c{m^2}$$

Solution

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